For the first hypergeometric function on the right-hand side we use [1]} with \(z=1\)
\(\,\mbox{}_{3}F_{2}\!\left(\genfrac{}{}{0.0pt}{}{a_1+1,a_2,a_3}{b_1,b_2};1\right)a_1(b_1+b_2-a_1-a_2-a_2-1)+ \\+\,\mbox{}_{3}F_{2}\!\left(\genfrac{}{}{0.0pt}{}{a_1,a_2,a_3}{b_1,b_2};1\right)\big ((2a_1-b_1)(2a_1-b_2)+a_1-(a_1)^2-(a_1-a_2)(a_1-a_3) \big )- \\-\,\mbox{}_{3}F_{2}\!\left(\genfrac{}{}{0.0pt}{}{a_1-1,a_2,a_3}{b_1,b_2};1\right)(a_1-b_1)(a_1-b_2)=0\) 
